If the expansion of  (1 + x)m(1 − x)n , the coefficients of  x  and  x2  are 3 and -6 respectively, then:

1. Break Down the Problem

We need to find the values of $m$ and $n$ such that in the expansion of $(1+x)^m (1-x)^n$, the coefficients of $x$ and $x^2$ equal 3 and -6, respectively.

2. Relevant Concepts

Using the Binomial Theorem, the coefficients of $x^k$ in $(1 + x)^m$ and $(1 - x)^n$ can be determined:

• The coefficient of $x$ in $(1+x)^m$ is $\binom{m}{1} = m$.

• The coefficient of $x$ in $(1-x)^n$ is $\binom{n}{1}(-1)^1 = -n$. Thus, the total coefficient of $x$ from the combined expansion is $m - n$.

• The coefficient of $x^2$ in $(1+x)^m$ is $\binom{m}{2} = \frac{m(m-1)}{2}$.

• The coefficient of $x^2$ in $(1-x)^n$ is $\binom{n}{2}(-1)^2 = \frac{n(n-1)}{2}$. Thus, the total coefficient of $x^2$ from the combined expansion is $\frac{m(m-1)}{2} + \frac{n(n-1)}{2}$.

3. Analysis and Detail

We now set up the equations based on the coefficients given in the problem:

1. For the coefficient of $x$: $m - n = 3 \hspace{30pt} \text{(Equation 1)}$

2. For the coefficient of $x^2$: $\frac{m(m-1)}{2} + \frac{n(n-1)}{2} = -6 \implies m(m-1) + n(n-1) = -12 \hspace{30pt} \text{(Equation 2)}$

4. Verify and Summarize

From Equation 1, we can express $m$ in terms of $n$: $m = n + 3$ Substituting $m = n + 3$ into Equation 2: $(n + 3)(n + 2) + n(n-1) = -12$ Expanding and simplifying: $(n^2 + 5n + 6) + (n^2 - n) = -12$ $2n^2 + 4n + 6 = -12$ $2n^2 + 4n + 18 = 0 \implies n^2 + 2n + 9 = 0$ The discriminant: $D = 2^2 - 4 \cdot 1 \cdot 9 = 4 - 36 = -32 < 0$ Thus there are no real solutions for $n$ and consequently for $m$ as well.

Final Answer

There are no real values of $m$ and $n$ that satisfy the given conditions for coefficients of $x$ and $x^2$ in the expansion of $(1 + x)^m(1 - x)^n$.